Question

The proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number resulting in a defect. Assume the births are independent.
1. Identify an appropriate probability model for X.
a. Uniform distribution with mean 2.5.
b. Poisson distribution with mean 5/33.
c. binomial distribution with n = 5 and p = 1/33.
d. binomial distribution with n = 5 and p = 32/33.
e. Normal distribution with mean 5 and variance 1/33.

Answer 1

Answer:

c. binomial distribution with n = 5 and p = 1/33.

Step-by-step explanation:

For each birth, there are only two possible outcomes. Either it results in a defect, or it does not. The probability of a birth resulting in a defect is independent of other births. So we use the binomial probability distrbution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The proportion of American births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC).

This means that the probability of a birth resulting in a defect is $p = \frac{1}{33}$

A local hospital randomly selects five births.

This means that $n = 5$

So the correct answer is:

c. binomial distribution with n = 5 and p = 1/33.